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3 years ago

Please help me on my homework! :) Its due tomorrow and I need help on numbers 3-6 thanks ;) (picture attached)

Mathematics

1 answer:

Aleks [24]3 years ago

7 0

3) 0.5(12m-22n)=0.5(12m)-0.5(22n)
=6m-11n
4) 2/3(18x+6z) =2/3(18x)-2/3(6z)
=12x-4z
5)2x+12=2(x+6)
6)12x+24=12(x+2)

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Pplllleeaaassseeeee hellppppp meeee<br> this is easy but i dont get it

AURORKA [14]

Answer:

choice 1) -6 and choice 2) -5

Step-by-step explanation:

choice 1)

-(1/2)-6)+6 > 8

9 > 8

choice 2)

-(1/2)(-5)+6 > 8

8 1/2 > 8

4 0

2 years ago

Solve for y when x = 5.<br> y = 3x + 4

pantera1 [17]

Answer:

Step-by-step explanation:

Plug in

y = 3(5) + 4
y = 19

5 0

2 years ago

Alessandro wrote the quadratic equation –6 = x2 + 4x – 1 in standard form. What is the value of c in his new equation?

Black_prince [1.1K]

Answer:

Step-by-step explanation:

To write it in standard form, we set the equation equal to 0. To do this, we add 6 to each side:

-6+6 = x² + 4x - 1 + 6

0 = x² + 4x + 5

The related function is

y = x² + 4x + 5

The value of c in this function is 5.

7 0

3 years ago

Cos^2x+cos^2(120°+x)+cos^2(120°-x)<br>i need this asap. pls help me

o-na [289]

Answer:

$\frac{3}{2}$

Step-by-step explanation:

Using the addition formulae for cosine

cos(x ± y) = cosxcosy ∓ sinxsiny

---------------------------------------------------------------

cos(120 + x) = cos120cosx - sin120sinx

= - cos60cosx - sin60sinx

= - $\frac{1}{2}$ cosx - $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos² (120 + x)

= $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------

cos(120 - x) = cos120cosx + sin120sinx

= -cos60cosx + sin60sinx

= - $\frac{1}{2}$ cosx + $\frac{\sqrt{3} }{2}$ sinx

squaring to obtain cos²(120 - x)

= $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

--------------------------------------------------------------------------

Putting it all together

cos²x + $\frac{1}{4}$ cos²x + $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x + $\frac{1}{4}$ cos²x - $\frac{\sqrt{3} }{2}$ sinxcosx + $\frac{3}{4}$ sin²x

= cos²x + $\frac{1}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ cos²x + $\frac{3}{2}$ sin²x

= $\frac{3}{2}$ (cos²x + sin²x) = $\frac{3}{2}$

5 0

3 years ago

Select the correct answer.

masya89 [10]

Answer:

Step-by-step explanation:

Either try all the numbers and see which ones work or do the following.

Times both sides by 2x-5.

3=x(2x-5)=2x^2-5x

Rearrange and solve the quadratic. I will factorise.

2x^2-5x-3=0

(2x+1)(x-3)=0

so x = -1/2 or 3

6 0

3 years ago